Some facts, better explained, from Atiyah-Macdonald

by ayushkhaitan3437

Today we shall discuss some interesting properties of elements of a ring.

1. If a\in R is not a unit, then it is present in some maximal ideal of the ring R. Self-explanatory.

2. If a is present in every maximal ideal, then 1+xa is a unit for all x\in R. Proof: Let 1+xa not be a unit. Then it is present in some maximal ideal (from 1). Let 1+xa=m, where m is an element from the maximal ideal 1+xa is a part of. Then 1=m-xa. Hence, 1 is also a member of the maximal ideal, which is absurd.

Let’s break down this theorem into elementary steps, and present a better proof (than given on pg.6 of “Commutative Algebra” by Atiyah-Macdonald). If x\in M_1 for some maximal ideal M_1, then 1\pm xy\notin M_1 for all y\in R. Similarly, If x\in M_2 for some maximal ideal M_2, then 1\pm xy\notin M_2 for all y\in R. This argument can then be extended to the fact that if x\in all maximal ideals, then 1\pm xy\notin any maximal ideal for all y\in R. An element not there in any maximal ideal is a unit. Hence, 1\pm xy is a unit.

3. If 1-xy is a unit for all y\in R, then x is part of every maximal ideal in R. Proof: Let is assume x is not part of some maximal ideal. Then there exists some m\in that maximal ideal such that m+xy=1. This implies that m=1-xy, which is impossible as 1-xy is a unit. The same argument can be used for 1+xy.

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On pg.6 of Atiyah-Macdonald, it is mentioned that if a,b,c are ideals, then a\cap (b+c)=a\cap b+a\cap c if b\subseteq a and c\subseteq a. It is not elaborated in the book, and the flippant reader may be confused. I’d like to elaborate on this concept.

a\cap (b+c) consists of those elements in b+c which are also there in a. Now elements of a\cap (b+c)=[(b+c) wrt those elements of both b and c that are in a] \bigcup [(b+c) wrt those elements of b that are in a and those elements of c that are not] \bigcup [(b+c) wrt those elements of b that are not in a and those elements of c that are] \bigcup [(b+c) wrt those elements of both b and c that are not in a]

The second and the third terms are null sets, as can be easily seen.

The fourth term is NOT necessarily empty. However, it becomes an empty set if b,c\subseteq a. It may also become an empty set under other conditions which ensure that if both b and c are not in a, then b+c\notin a.

In summation, a\cap (b+c)=a\cap b+a\cap c is definitely true when b,c\subseteq a. However, it is also true under other conditions which ensure that the fourth term is an empty set.

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I have extended a small paragraph in Atiyah-Macdonald to a full-fledged exposition: (a,b,c and d are ideals in commutative ring R)

1. a\cup (b+c)\subseteq(a\cup b)+(a\cup c)– Both sides contain all elements of a and b+c. Remember that b\cup c\subseteq b+c. However, the right hand side also contains elements of the form a+b and a+c, which the left hand side does not contain.

2. a\cap (b+c)– This has already been explained above.

3. a+(b\cup c)=(a+b)\cup (a+c)– Both are exactly the same.

4. a+ (b\cap c)\subseteq (a+b)\cap (a+c)– There might be b_1,c_1\notin b\cap c such that a'+b_1=a''+c_1. However, any element in a+(b\cap c) will definitely be present in (a+ b)\cap (a+c).

5. a(b\cup c)\supseteq ab\cup ac– LHS contains elements of the form a'b_1+a''c_1, which the RHS doesn’t. In fact, LHS is an ideal while the RHS isn’t. You might wonder how LHS is an ideal. I have just extended the algorithm used to make AB an ideal when A and B are both ideals to situations in which A is an ideal and B is any subset of R.

6. a(b\cap c)\subseteq ab\cap ac– The RHS may contain elements of the form a'b_1=a''c_1 for b_1,c_1\notin b\cap c.

7. a(b+c)=ab+ac– Easy enough to see.

8. (a+b)(c\cap d)=(c\cap d)a+(c\cap d)b\subseteq(ac\cap ad)+(bc\cap bd)

From this formula, we have (a+b)(a\cap b)\subseteq (a\cap ab)+(b\cap ab)\subseteq ab.
This fact is mentioned on pg.7 of Atiyah-Macdonald.

9. (a+b)(c\cup d)= (c\cup d)a+(c\cup d)b= (ca\cup da)+(cb\cup db).

From this formula, we have (a+b)(a\cup b)=(a^2\cup ab)+(b^2\cup ab)\supseteq ab.

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